Practice MCQ Questions and Answer on Area

Solution:

 $$\eqalign{ & 2\pi R = 100 \cr & R = \frac{{100}}{{2\pi }} = \frac{{50}}{\pi } \cr & R = \frac{1}{2} \times {\text{diagonal}} \cr & \Rightarrow {\text{Diagonal}} = 2R \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2 \times 50}}{\pi } \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{100}}{\pi } \cr} $$ $$\eqalign{ & \therefore {\text{Area of the square}} = \frac{1}{2} \times {\left( {{\text{diagonal}}} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{a^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2} \times {\left( {\frac{{100}}{\pi }} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{\sqrt 2 }} \times \frac{{100}}{\pi } \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50\sqrt 2 }}{\pi }cm \cr} $$

Solution:

 $$\eqalign{ & \frac{l}{{2\left( {l + b} \right)}} = \frac{1}{3} \cr & \Rightarrow 3l = 2l + 2b \cr & \Rightarrow l = 2b \cr & \Rightarrow \frac{1}{b} = \frac{2}{1} = 2:1 \cr} $$

Solution:

 Let the length of the rectangle be x metres. Then, breath of the rectangle = $$\left( {\frac{{168}}{x}} \right)m$$$$\eqalign{ & \therefore \sqrt {{x^2} + {{\left( {\frac{{168}}{x}} \right)}^2}} = 25 \cr & \Rightarrow \sqrt {{x^2} + \frac{{28224}}{{{x^2}}}} = 25 \cr & \Rightarrow {x^2} + \frac{{28224}}{{{x^2}}} = 625 \cr & \Rightarrow {x^4} - 625{x^2} + 28224 = 0 \cr & \Rightarrow {x^4} - 576{x^2} - 49{x^2} + 28224 = 0 \cr & \Rightarrow {x^2}\left( {{x^2} - 576} \right) - 49\left( {{x^2} - 576} \right) = 0 \cr & \Rightarrow \left( {{x^2} - 576} \right)\left( {{x^2} - 49} \right) = 0 \cr & \Rightarrow {x^2} = 576\,\,or\,\,{x^2} = 49 \cr & \Rightarrow x = 24\,\,\,or\,\,\,x = 7 \cr} $$ Hence, length = 24 m and breadth = 7 m

Solution:

 Let the breadth = x metres and length = (2x) metres Area of 4 walls = [2(2x + x)× 4] m2 = (24x) m2 ∴ 24x = 120 ⇒ x = 5 So, length = 10 m, and breadth = 5 m Area of the floor = (10 × 5) m2 = 50 m2

Solution:

 Let the breadth and height of the room be b metres and h metres respectively. Then, length of the room = (2b) metres Area of the ceiling : $$\eqalign{ & = \left( {2b \times b} \right)m \cr & = \left( {2{b^2}} \right){m^2} \cr} $$ $$\eqalign{ & 2{b^2} = \frac{{5000}}{{25}} \cr & 2{b^2} = 200 \cr & {b^2} = 100 \cr & b = 10 \cr} $$ So, length = 20 m, breadth = 10 m Area of 4 walls : $$\eqalign{ & = \left[ {2\left( {20 + 10} \right) \times h} \right]{m^2} \cr & = \left( {60h} \right){m^2} \cr} $$ $$\eqalign{ & \therefore 60h = \frac{{64800}}{{240}} \cr & \Rightarrow 60h = 270 \cr & \Rightarrow h = \frac{{270}}{{60}} \cr & \Rightarrow h = 4.5\,m \cr} $$

Solution:

 Area left after laying black tiles : $$\eqalign{ & = \left[ {\left( {20 - 4} \right) \times \left( {10 - 4} \right)} \right]{\text{sq}}{\text{.ft.}} \cr & = 96\,{\text{sq}}{\text{.ft.}} \cr} $$ Area under white tiles : $$\eqalign{ & = \left( {\frac{1}{3} \times 96} \right){\text{sq}}{\text{.ft.}} \cr & = 32\,{\text{sq}}{\text{.ft.}} \cr} $$ Area under blue tiles : $$\eqalign{ & = \left( {96 - 32} \right){\text{sq}}{\text{.ft.}} \cr & = 64\,{\text{sq}}{\text{.ft.}} \cr} $$ Number of blue tiles : $$\eqalign{ & = \frac{{64}}{{\left( {2 \times 2} \right)}} \cr & = 16 \cr} $$

Solution:

 Let original length = x and original breadth = y Then, original area = xy New length : $$\eqalign{ & = \frac{{160x}}{{100}} \cr & = \frac{{8x}}{5} \cr} $$ Let the new breadth = z Then, $$\eqalign{ & \Rightarrow \frac{{8x}}{5} \times z = xy \cr & \Rightarrow z = \frac{{5y}}{8} \cr} $$ ∴ Decrease in breadth : $$\eqalign{ & = \left( {\frac{{3y}}{8} \times \frac{1}{y} \times 100} \right)\% \cr & = 37\frac{1}{2}\% \cr} $$

Solution:

 Let the radius of circle be r cm and side of square be a cm Then circumference of circle = $$2\pi r$$ and Perimeter of square = 4a According to the question, $$\eqalign{ & 2\pi r = 4a \times \frac{{110}}{{100}} \cr & \Rightarrow 2\pi r = \frac{{44a}}{{10}} \cr & \Rightarrow r = \frac{{44a}}{{2\pi \times 10}} \cr & \Rightarrow r = \frac{{11a}}{{5\pi }} \cr & \Rightarrow a = \frac{{5\pi r}}{{11 }} . . . . .(i) \cr & {\text{Also, }}\pi {r^2} - {a^2} \Rightarrow 216 \cr} $$ $$ \Rightarrow \pi {r^2} - \frac{{25\pi {r^2}}}{{121}} = 216$$     $$\left[ {{\text{from equation (i)}}} \right]$$ $$\eqalign{ & \Rightarrow \frac{{121\pi {r^2} - 25\pi {r^2}}}{{121}} = 216 \cr & \Rightarrow {r^2}\left[ {121\pi - 25{\pi ^2}} \right] = 26136 \cr} $$ $$ \Rightarrow {r^2}\left[ {121 \times \frac{{22}}{7} - 25 \times \frac{{22}}{7} \times \frac{{22}}{7}} \right]$$      $$ = 26136$$ $$\eqalign{ & \Rightarrow {r^2}\left[ {\frac{{2662}}{7} - \frac{{12100}}{{49}}} \right] = 26136 \cr & \Rightarrow {r^2}\left[ {\frac{{6534}}{{49}}} \right] = 26136 \cr & \Rightarrow {r^2} = \frac{{26136 \times 49}}{{6534}} \cr & \Rightarrow {r^2} = 196 \cr & \Rightarrow r = 14\,cm \cr & \therefore a = \frac{{5\pi r}}{{11}} = 5 \times \frac{{22}}{7} \times \frac{{14}}{{11}} = 20\,cm \cr} $$ Hence, diagonal of square $$ = \sqrt 2 a = 20\sqrt 2 \,cm$$

Solution:

 Let l and b be the length and breadth of the rectangle respectively. Then, $$\eqalign{ & \Rightarrow \sqrt {{l^2} + {b^2}} = 15 \cr & \Rightarrow \left( {{l^2} + {b^2}} \right) = {\left( {15} \right)^2} \cr & \Rightarrow {l^2} + {b^2} = 225 \cr} $$ And, $$\eqalign{ & \Rightarrow l + b = 3 \cr & \Rightarrow {\left( {l - b} \right)^2} = 9 \cr & \Rightarrow {l^2} + {b^2} - 2lb = 9 \cr & \Rightarrow 225 - 2lb = 9 \cr & \Rightarrow 2lb = 216 \cr & \Rightarrow lb = 108 \cr} $$ Hence, area of the field $$ = lb = 108\,{m^2}$$

Solution:

 Let the sides of the triangles be x cm, (x + 1) cm and (x + 2) cm respectively. Then, $$\eqalign{ & x + \left( {x + 1} \right) + \left( {x + 2} \right) = 120 \cr & \Rightarrow 3x + 3 = 120 \cr & \Rightarrow 3x = 117 \cr & \Rightarrow x = 39 \cr} $$ ∴ Length of greatest side : = (39 + 2) cm = 41 cm